Integrand size = 66, antiderivative size = 22 \[ \int \frac {e^{e^{\frac {e^x \left (-1+e^4+20 x\right )}{5 x}}+x+\frac {e^x \left (-1+e^4+20 x\right )}{5 x}} \left (1+e^4 (-1+x)-x+20 x^2\right )}{5 x^2} \, dx=e^{e^{e^x \left (4+\frac {-1+e^4}{5 x}\right )}} \] Output:
exp(exp(exp(x)*(1/5*(exp(4)-1)/x+4)))
Time = 5.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {e^{e^{\frac {e^x \left (-1+e^4+20 x\right )}{5 x}}+x+\frac {e^x \left (-1+e^4+20 x\right )}{5 x}} \left (1+e^4 (-1+x)-x+20 x^2\right )}{5 x^2} \, dx=e^{e^{\frac {e^x \left (-1+e^4+20 x\right )}{5 x}}} \] Input:
Integrate[(E^(E^((E^x*(-1 + E^4 + 20*x))/(5*x)) + x + (E^x*(-1 + E^4 + 20* x))/(5*x))*(1 + E^4*(-1 + x) - x + 20*x^2))/(5*x^2),x]
Output:
E^E^((E^x*(-1 + E^4 + 20*x))/(5*x))
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (20 x^2-x+e^4 (x-1)+1\right ) \exp \left (x+e^{\frac {e^x \left (20 x+e^4-1\right )}{5 x}}+\frac {e^x \left (20 x+e^4-1\right )}{5 x}\right )}{5 x^2} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int \frac {\exp \left (-\frac {e^x \left (-20 x-e^4+1\right )}{5 x}+e^{-\frac {e^x \left (-20 x-e^4+1\right )}{5 x}}+x\right ) \left (20 x^2-x-e^4 (1-x)+1\right )}{x^2}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \frac {1}{5} \int \frac {\exp \left (-\frac {e^x \left (-20 x-e^4+1\right )}{5 x}+e^{-\frac {e^x \left (-20 x-e^4+1\right )}{5 x}}+x\right ) \left (20 x^2-\left (1-e^4\right ) x-e^4+1\right )}{x^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{5} \int \left (20 \exp \left (-\frac {e^x \left (-20 x-e^4+1\right )}{5 x}+e^{-\frac {e^x \left (-20 x-e^4+1\right )}{5 x}}+x\right )+\frac {\exp \left (-\frac {e^x \left (-20 x-e^4+1\right )}{5 x}+e^{-\frac {e^x \left (-20 x-e^4+1\right )}{5 x}}+x\right ) \left (-1+e^4\right )}{x}+\frac {\exp \left (-\frac {e^x \left (-20 x-e^4+1\right )}{5 x}+e^{-\frac {e^x \left (-20 x-e^4+1\right )}{5 x}}+x\right ) \left (1-e^4\right )}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (\left (1-e^4\right ) \int \frac {\exp \left (-\frac {e^x \left (-20 x-e^4+1\right )}{5 x}+e^{-\frac {e^x \left (-20 x-e^4+1\right )}{5 x}}+x\right )}{x^2}dx+20 \int \exp \left (-\frac {e^x \left (-20 x-e^4+1\right )}{5 x}+e^{-\frac {e^x \left (-20 x-e^4+1\right )}{5 x}}+x\right )dx-\left (1-e^4\right ) \int \frac {\exp \left (-\frac {e^x \left (-20 x-e^4+1\right )}{5 x}+e^{-\frac {e^x \left (-20 x-e^4+1\right )}{5 x}}+x\right )}{x}dx\right )\) |
Input:
Int[(E^(E^((E^x*(-1 + E^4 + 20*x))/(5*x)) + x + (E^x*(-1 + E^4 + 20*x))/(5 *x))*(1 + E^4*(-1 + x) - x + 20*x^2))/(5*x^2),x]
Output:
$Aborted
Time = 5.12 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77
method | result | size |
norman | \({\mathrm e}^{{\mathrm e}^{\frac {\left ({\mathrm e}^{4}+20 x -1\right ) {\mathrm e}^{x}}{5 x}}}\) | \(17\) |
risch | \({\mathrm e}^{{\mathrm e}^{\frac {\left ({\mathrm e}^{4}+20 x -1\right ) {\mathrm e}^{x}}{5 x}}}\) | \(17\) |
parallelrisch | \({\mathrm e}^{{\mathrm e}^{\frac {\left ({\mathrm e}^{4}+20 x -1\right ) {\mathrm e}^{x}}{5 x}}}\) | \(17\) |
Input:
int(1/5*((-1+x)*exp(4)+20*x^2-x+1)*exp(x)*exp(1/5*(exp(4)+20*x-1)*exp(x)/x )*exp(exp(1/5*(exp(4)+20*x-1)*exp(x)/x))/x^2,x,method=_RETURNVERBOSE)
Output:
exp(exp(1/5*(exp(4)+20*x-1)*exp(x)/x))
Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (16) = 32\).
Time = 0.11 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.64 \[ \int \frac {e^{e^{\frac {e^x \left (-1+e^4+20 x\right )}{5 x}}+x+\frac {e^x \left (-1+e^4+20 x\right )}{5 x}} \left (1+e^4 (-1+x)-x+20 x^2\right )}{5 x^2} \, dx=e^{\left (-x - \frac {{\left (20 \, x + e^{4} - 1\right )} e^{x}}{5 \, x} + \frac {5 \, x^{2} + {\left (20 \, x + e^{4} - 1\right )} e^{x} + 5 \, x e^{\left (\frac {{\left (20 \, x + e^{4} - 1\right )} e^{x}}{5 \, x}\right )}}{5 \, x}\right )} \] Input:
integrate(1/5*((-1+x)*exp(4)+20*x^2-x+1)*exp(x)*exp(1/5*(exp(4)+20*x-1)*ex p(x)/x)*exp(exp(1/5*(exp(4)+20*x-1)*exp(x)/x))/x^2,x, algorithm="fricas")
Output:
e^(-x - 1/5*(20*x + e^4 - 1)*e^x/x + 1/5*(5*x^2 + (20*x + e^4 - 1)*e^x + 5 *x*e^(1/5*(20*x + e^4 - 1)*e^x/x))/x)
Time = 0.58 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {e^{e^{\frac {e^x \left (-1+e^4+20 x\right )}{5 x}}+x+\frac {e^x \left (-1+e^4+20 x\right )}{5 x}} \left (1+e^4 (-1+x)-x+20 x^2\right )}{5 x^2} \, dx=e^{e^{\frac {\left (4 x - \frac {1}{5} + \frac {e^{4}}{5}\right ) e^{x}}{x}}} \] Input:
integrate(1/5*((-1+x)*exp(4)+20*x**2-x+1)*exp(x)*exp(1/5*(exp(4)+20*x-1)*e xp(x)/x)*exp(exp(1/5*(exp(4)+20*x-1)*exp(x)/x))/x**2,x)
Output:
exp(exp((4*x - 1/5 + exp(4)/5)*exp(x)/x))
Time = 0.38 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {e^{e^{\frac {e^x \left (-1+e^4+20 x\right )}{5 x}}+x+\frac {e^x \left (-1+e^4+20 x\right )}{5 x}} \left (1+e^4 (-1+x)-x+20 x^2\right )}{5 x^2} \, dx=e^{\left (e^{\left (\frac {e^{\left (x + 4\right )}}{5 \, x} - \frac {e^{x}}{5 \, x} + 4 \, e^{x}\right )}\right )} \] Input:
integrate(1/5*((-1+x)*exp(4)+20*x^2-x+1)*exp(x)*exp(1/5*(exp(4)+20*x-1)*ex p(x)/x)*exp(exp(1/5*(exp(4)+20*x-1)*exp(x)/x))/x^2,x, algorithm="maxima")
Output:
e^(e^(1/5*e^(x + 4)/x - 1/5*e^x/x + 4*e^x))
\[ \int \frac {e^{e^{\frac {e^x \left (-1+e^4+20 x\right )}{5 x}}+x+\frac {e^x \left (-1+e^4+20 x\right )}{5 x}} \left (1+e^4 (-1+x)-x+20 x^2\right )}{5 x^2} \, dx=\int { \frac {{\left (20 \, x^{2} + {\left (x - 1\right )} e^{4} - x + 1\right )} e^{\left (x + \frac {{\left (20 \, x + e^{4} - 1\right )} e^{x}}{5 \, x} + e^{\left (\frac {{\left (20 \, x + e^{4} - 1\right )} e^{x}}{5 \, x}\right )}\right )}}{5 \, x^{2}} \,d x } \] Input:
integrate(1/5*((-1+x)*exp(4)+20*x^2-x+1)*exp(x)*exp(1/5*(exp(4)+20*x-1)*ex p(x)/x)*exp(exp(1/5*(exp(4)+20*x-1)*exp(x)/x))/x^2,x, algorithm="giac")
Output:
integrate(1/5*(20*x^2 + (x - 1)*e^4 - x + 1)*e^(x + 1/5*(20*x + e^4 - 1)*e ^x/x + e^(1/5*(20*x + e^4 - 1)*e^x/x))/x^2, x)
Time = 7.98 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {e^{e^{\frac {e^x \left (-1+e^4+20 x\right )}{5 x}}+x+\frac {e^x \left (-1+e^4+20 x\right )}{5 x}} \left (1+e^4 (-1+x)-x+20 x^2\right )}{5 x^2} \, dx={\mathrm {e}}^{{\mathrm {e}}^{-\frac {{\mathrm {e}}^x}{5\,x}}\,{\mathrm {e}}^{4\,{\mathrm {e}}^x}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^4\,{\mathrm {e}}^x}{5\,x}}} \] Input:
int((exp((exp(x)*(20*x + exp(4) - 1))/(5*x))*exp(exp((exp(x)*(20*x + exp(4 ) - 1))/(5*x)))*exp(x)*(exp(4)*(x - 1) - x + 20*x^2 + 1))/(5*x^2),x)
Output:
exp(exp(-exp(x)/(5*x))*exp(4*exp(x))*exp((exp(4)*exp(x))/(5*x)))
Time = 0.19 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.64 \[ \int \frac {e^{e^{\frac {e^x \left (-1+e^4+20 x\right )}{5 x}}+x+\frac {e^x \left (-1+e^4+20 x\right )}{5 x}} \left (1+e^4 (-1+x)-x+20 x^2\right )}{5 x^2} \, dx=e^{\frac {e^{\frac {e^{x} e^{4}+20 e^{x} x}{5 x}}}{e^{\frac {e^{x}}{5 x}}}} \] Input:
int(1/5*((-1+x)*exp(4)+20*x^2-x+1)*exp(x)*exp(1/5*(exp(4)+20*x-1)*exp(x)/x )*exp(exp(1/5*(exp(4)+20*x-1)*exp(x)/x))/x^2,x)
Output:
e**(e**((e**x*e**4 + 20*e**x*x)/(5*x))/e**(e**x/(5*x)))